• # question_answer For what value of $\lambda$the sum of the squares of the roots of ${{x}^{2}}+(2+\lambda )\,x-\frac{1}{2}(1+\lambda )=0$ is minimum   [AMU 1999] A) 3/2 B) 1 C) 1/2 D) 11/4

Given equation is ${{x}^{2}}+(2+\lambda )x-\frac{1}{2}(1+\lambda )=0$ So, $\alpha +\beta =-(2+\lambda )=0$ and $\alpha \beta =-\left( \frac{1+\lambda }{2} \right)$ Now, ${{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta$ $\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}={{\left[ -(2+\lambda ) \right]}^{2}}+2.\frac{(1+\lambda )}{2}$ Þ ${{\alpha }^{2}}+{{\beta }^{2}}={{\lambda }^{2}}+4+4\lambda +1+\lambda ={{\lambda }^{2}}+5\lambda +5$ which is minimum for $\lambda =1/2$.